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After searching for a while, I realized there would be no simple expression for this integral. I found the following expressions in Abramowitz and Stegun, which you can download. They involve confluent hypergeometric functions $M(a,b,z)$ for the most general one.

There are some conditions on the range of values for the parameters for these formulae to hold, but I think that should not be a problem in your case. You can find them in the reference as well as further details on confluent hypergeometric functions.

Also I did some experiments in Mathematica and it seems J.M. may be right: $\nu = 0: -\frac{1}{16}e^{-\frac{r^2}{4}}\sqrt{\pi}(-6 + r^2)$, $\nu = 1: \frac{r(r^2-2)+(4 + 4r^2 - r^4)DawsonF(r/2)}{8r^2}$ For larger $\nu$ the pattern is this: even $\nu$: $e^{-\frac{r^2}{4}}$ x a polynominal fraction of r + possibly a term with error function(r/2), for odd $\nu$: a polynominal in r x a Dawson integral(r/2). However I only tested up to $\nu=6$ so I do not know if this approach is generally applicable.
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AndyApr 25 '11 at 11:51

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Thank you J.M! Using Mathematica I should be able to evaluate a few Ms and then use the recurrence relations in Abramowitz and Stegun to calculate all other Ms.
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AndyApr 25 '11 at 12:11